Answer :
To determine if an ordered pair [tex]\((x, y)\)[/tex] is a solution to the function [tex]\( f(x) = 3 - 2x \)[/tex], we need to check if substituting [tex]\( x \)[/tex] into the function [tex]\( f(x) \)[/tex] results in the value [tex]\( y \)[/tex].
Let's evaluate each of the given ordered pairs against the function [tex]\( f(x) = 3 - 2x \)[/tex]:
1. [tex]\((-2, -1)\)[/tex]:
- Substitute [tex]\( x = -2 \)[/tex] into [tex]\( f(x) \)[/tex]:
[tex]\[ f(-2) = 3 - 2(-2) = 3 + 4 = 7 \][/tex]
- The result is [tex]\( 7 \)[/tex], not [tex]\(-1\)[/tex]. Therefore, [tex]\((-2, -1)\)[/tex] is not a solution to the function.
2. [tex]\((1, 0)\)[/tex]:
- Substitute [tex]\( x = 1 \)[/tex] into [tex]\( f(x) \)[/tex]:
[tex]\[ f(1) = 3 - 2(1) = 3 - 2 = 1 \][/tex]
- The result is [tex]\( 1 \)[/tex], not [tex]\( 0\)[/tex]. Therefore, [tex]\((1, 0)\)[/tex] is not a solution to the function.
3. [tex]\((2, -1)\)[/tex]:
- Substitute [tex]\( x = 2 \)[/tex] into [tex]\( f(x) \)[/tex]:
[tex]\[ f(2) = 3 - 2(2) = 3 - 4 = -1 \][/tex]
- The result is [tex]\(-1\)[/tex], which matches [tex]\( y \)[/tex]. Therefore, [tex]\((2, -1)\)[/tex] is a solution to the function.
4. [tex]\((0, 3)\)[/tex]:
- Substitute [tex]\( x = 0 \)[/tex] into [tex]\( f(x) \)[/tex]:
[tex]\[ f(0) = 3 - 2(0) = 3 - 0 = 3 \][/tex]
- The result is [tex]\( 3 \)[/tex], which matches [tex]\( y \)[/tex]. Therefore, [tex]\((0, 3)\)[/tex] is a solution to the function.
5. [tex]\((-1, 5)\)[/tex]:
- Substitute [tex]\( x = -1 \)[/tex] into [tex]\( f(x) \)[/tex]:
[tex]\[ f(-1) = 3 - 2(-1) = 3 + 2 = 5 \][/tex]
- The result is [tex]\( 5 \)[/tex], which matches [tex]\( y \)[/tex]. Therefore, [tex]\((-1, 5)\)[/tex] is a solution to the function.
Hence, the ordered pairs that are solutions to the function [tex]\( f(x) = 3 - 2x \)[/tex] are:
- [tex]\((2, -1)\)[/tex]
- [tex]\((0, 3)\)[/tex]
- [tex]\((-1, 5)\)[/tex]
The selected ordered pairs that satisfy the function are [tex]\((2, -1)\)[/tex], [tex]\((0, 3)\)[/tex], and [tex]\((-1, 5)\)[/tex].
Let's evaluate each of the given ordered pairs against the function [tex]\( f(x) = 3 - 2x \)[/tex]:
1. [tex]\((-2, -1)\)[/tex]:
- Substitute [tex]\( x = -2 \)[/tex] into [tex]\( f(x) \)[/tex]:
[tex]\[ f(-2) = 3 - 2(-2) = 3 + 4 = 7 \][/tex]
- The result is [tex]\( 7 \)[/tex], not [tex]\(-1\)[/tex]. Therefore, [tex]\((-2, -1)\)[/tex] is not a solution to the function.
2. [tex]\((1, 0)\)[/tex]:
- Substitute [tex]\( x = 1 \)[/tex] into [tex]\( f(x) \)[/tex]:
[tex]\[ f(1) = 3 - 2(1) = 3 - 2 = 1 \][/tex]
- The result is [tex]\( 1 \)[/tex], not [tex]\( 0\)[/tex]. Therefore, [tex]\((1, 0)\)[/tex] is not a solution to the function.
3. [tex]\((2, -1)\)[/tex]:
- Substitute [tex]\( x = 2 \)[/tex] into [tex]\( f(x) \)[/tex]:
[tex]\[ f(2) = 3 - 2(2) = 3 - 4 = -1 \][/tex]
- The result is [tex]\(-1\)[/tex], which matches [tex]\( y \)[/tex]. Therefore, [tex]\((2, -1)\)[/tex] is a solution to the function.
4. [tex]\((0, 3)\)[/tex]:
- Substitute [tex]\( x = 0 \)[/tex] into [tex]\( f(x) \)[/tex]:
[tex]\[ f(0) = 3 - 2(0) = 3 - 0 = 3 \][/tex]
- The result is [tex]\( 3 \)[/tex], which matches [tex]\( y \)[/tex]. Therefore, [tex]\((0, 3)\)[/tex] is a solution to the function.
5. [tex]\((-1, 5)\)[/tex]:
- Substitute [tex]\( x = -1 \)[/tex] into [tex]\( f(x) \)[/tex]:
[tex]\[ f(-1) = 3 - 2(-1) = 3 + 2 = 5 \][/tex]
- The result is [tex]\( 5 \)[/tex], which matches [tex]\( y \)[/tex]. Therefore, [tex]\((-1, 5)\)[/tex] is a solution to the function.
Hence, the ordered pairs that are solutions to the function [tex]\( f(x) = 3 - 2x \)[/tex] are:
- [tex]\((2, -1)\)[/tex]
- [tex]\((0, 3)\)[/tex]
- [tex]\((-1, 5)\)[/tex]
The selected ordered pairs that satisfy the function are [tex]\((2, -1)\)[/tex], [tex]\((0, 3)\)[/tex], and [tex]\((-1, 5)\)[/tex].